# Calculus

## Integration of Functions # Area of a Region Bounded by Curves

• ### Area in Rectangular Coordinates

Recall that the area under the graph of a continuous function $$f\left( x \right)$$ between the vertical lines $$x = a,$$ $$x = b$$ can be computed by the definite integral:

$A = \int\limits_a^b {f\left( x \right)dx} = F\left( b \right) – F\left( a \right),$

where $$F\left( x \right)$$ is any antiderivative of $$f\left( x \right).$$

We can extend the notion of the area under a curve and consider the area of the region between two curves.

If $$f\left( x \right)$$ and $$g\left( x \right)$$ are two continuous functions and $$f\left( x \right) \ge g\left( x \right)$$ on the closed interval $$\left[ {a,b} \right],$$ then the area between the curves $$y = f\left( x \right)$$ and $$y = g\left( x \right)$$ in this interval is given by

$A = \int\limits_a^b {\left[ {f\left( x \right) – g\left( x \right)} \right]dx}.$

In terms of antiderivatives, the area of region is expressed in the form

${A = \int\limits_a^b {\left[ {f\left( x \right) – g\left( x \right)} \right]dx} }={ F\left( b \right) – G\left( b \right) – F\left( a \right) + G\left( a \right),}$

where $$F\left( x \right)$$ and $$G\left( x \right)$$ are antiderivatives of the functions $$f\left( x \right)$$ and $$g\left( x \right),$$ respectively.

Note that this area will always be non-negative as $$f\left( x \right) – g\left( x \right) \ge 0$$ for all $$x \in \left[ {a,b} \right].$$

If there are intersection points, we should break up the interval into several subintervals and determine which curve is greater on each subinterval. Then we can determine the area of each region by integrating the difference of the larger and the smaller function.

### Area in Polar Coordinates

Consider the region $$OKM$$ bounded by a polar curve $$r = f\left( \theta \right)$$ and two semi-straight lines $$\theta =\alpha$$ and $$\theta = \beta.$$

The area of the polar region is given by

${A = \frac{1}{2}\int\limits_\alpha ^\beta {{r^2}d\theta } }={ \frac{1}{2}\int\limits_\alpha ^\beta {{f^2}\left( \theta \right)d\theta } .}$

The area of a region between two polar curves $$r = f\left( \theta \right)$$ and $$r = g\left( \theta \right)$$ in the sector $$\left[ {\alpha ,\beta } \right]$$ is expressed by the integral

$A = \frac{1}{2}\int\limits_\alpha ^\beta {\left[ {{f^2}\left( \theta \right) – {g^2}\left( \theta \right)} \right]d\theta } .$

### Area of a Region Bounded by a Parametric Curve

Recall that the area under a curve $$y = f\left( x \right)$$ for $$f\left( x \right) \ge 0$$ on the interval $$\left[ {a,b} \right]$$ can be computed with the integral $$\int\limits_a^b {f\left( x \right)dx}.$$ Suppose now that the curve is defined in parametric form by the equations

${x = x\left( t \right),\;}\kern0pt{y = y\left( t \right).}$

If the parameter $$t$$ runs between $${t_1}$$ and $${t_2}$$ where

${a = x\left( {{t_1}} \right),\;}\kern0pt{b = x\left( {{t_2}} \right),}$

then the area under the curve is given by the formula

${A = \int\limits_a^b {f\left( x \right)dx} }={ \int\limits_a^b {ydx} }={ \int\limits_{{t_1}}^{{t_2}} {y\left( t \right)dx\left( t \right)} }={ \int\limits_{{t_1}}^{{t_2}} {y\left( t \right)dx\left( t \right)} }={ \int\limits_{{t_1}}^{{t_2}} {y\left( t \right)x^\prime\left( t \right)dt} .}$

The functions $$x\left( t \right),$$ $$x^\prime\left( t \right),$$ $$y\left( t \right)$$ here are assumed to be continuous on the interval $$\left[ {a,b} \right].$$ Besides that, the function $$x\left( t \right),$$ must be monotonic on this interval.

If $$x = x\left( t \right),$$ $$y = y\left( t \right),$$ $$0 \le t \le T$$ are parametric equations of a smooth piecewise closed curve $$C$$ traversed in the counterclockwise direction and bounding a region on the left (Figure $$5$$), then the area of the region is given by the following integrals:

${A = – \int\limits_0^T {y\left( t \right)x^\prime\left( t \right)dt} }={ \int\limits_0^T {x\left( t \right)y^\prime\left( t \right)dt} }={ \frac{1}{2}\int\limits_0^T {\left[ {x\left( t \right)y^\prime\left( t \right) – x^\prime\left( t \right)y\left( t \right)} \right]dt} .}$

• ## Solved Problems

Click a problem to see the solution.

### Example 1

Find the area between the curves $$y = {x^3}$$ and $$y = 3x + 2.$$

### Example 2

At what value of the parameter $$b\left( {b \gt 1} \right)$$ the area under the curve $$y = {x^2}$$ on the interval $$\left[{1,b}\right]$$ is equal to $$1?$$

### Example 3

Find the coordinate of the point $$a$$ that splits the area under the root function $$y = \sqrt{x}$$ on the interval $$\left[{0,4}\right]$$ into equal parts.

### Example 4

The region is bounded by the vertical lines $$x = t$$, $$x = t + \large{\frac{\pi }{2}}\normalsize$$, the $$x-$$axis, and the curve $$y = a + \cos x,$$ where $$a \ge 1.$$ Determine the value of $$t$$ at which the region has the largest area.

### Example 5

Find the area of the region enclosed by the curve $$y = \sqrt {x + 1}$$ and the line $$y = x + 1.$$

### Example 6

Find the area of the region enclosed by the root curve $$y = \sqrt {x}$$ and the line $$y = kx,$$ where $$k \gt 0.$$

### Example 7

Find the area of the region bounded by the curve $$y = {2^x}$$ and the lines $$x = 0,$$ $$y = 2.$$

### Example 8

Find the area enclosed by the three petaled rose $$r = \sin 3\theta .$$

### Example 9

Find the area enclosed by the cardioid $$r = 1 + \cos \theta .$$

### Example 10

Find the area of the region bounded by the astroid $${x^{\frac{3}{2}}} + {y^{\frac{3}{2}}} = 1.$$

### Example 1.

Find the area between the curves $$y = {x^3}$$ and $$y = 3x + 2.$$

Solution.

First we determine the points of intersection of the curves.

We set $$f\left( x \right) = g\left( x \right)$$ to find the roots:

${{x^3} = 3x + 2,}\;\; \Rightarrow {{x^3} – 3x – 2 = 0.}$

$\Rightarrow {x^3} + \underbrace {{x^2} – {x^2}}_0 – 3x – 2 = 0,$

$\Rightarrow {x^3} + {x^2} – {x^2} – x – 2x – 2 = 0,$

${\Rightarrow {x^2}\left( {x + 1} \right) – x\left( {x + 1} \right) }-{ 2\left( {x + 1} \right) }={ 0,}$

$\Rightarrow \left( {x + 1} \right)\left( {{x^2} – x – 2} \right) = 0.$

${{x^2} – x – 2 = 0,}\;\; \Rightarrow {D = {\left( { – 1} \right)^2} + 4 \cdot \left( { – 2} \right) = 9,}\;\; \Rightarrow {{x_{1,2}} = \frac{{1 \pm \sqrt 9 }}{2} = 2, – 1.}$

Thus, the cubic equation has two roots: $$x = -1$$ (of multiplicity $$2$$) and $$x=2.$$

The area we wish to calculate is shown in Figure $$6$$ below.

We can see from the Figure that the line $$y = 3x + 2$$ lies above the cubic parabola $$y = {x^3}$$ on the interval $$\left[ { – 1,2} \right].$$ Hence, the area of the region is given by

$\require{cancel}{A = \int\limits_{ – 1}^2 {\left[ {\left( {3x + 2} \right) – {x^3}} \right]dx} }={ \int\limits_{ – 1}^2 {\left( {3x + 2 – {x^3}} \right)dx} }={ \left. {\frac{{3{x^2}}}{2} + 2x – \frac{{{x^4}}}{4}} \right|_{ – 1}^2 }={ \left( {6 + \cancel{4} – \cancel{4}} \right) }-{ \left( {\frac{3}{2} – 2 – \frac{1}{4}} \right) }={ \frac{{27}}{4}.}$

### Example 2.

At what value of the parameter $$b\left( {b \gt 1} \right)$$ the area under the curve $$y = {x^2}$$ on the interval $$\left[{1,b}\right]$$ is equal to $$1?$$

Solution.

The area under the curve is given by the integral

$A = \int\limits_1^b {{x^2}dx} = 1.$

Integrating yields:

${\int\limits_1^b {{x^2}dx} = \left. {\frac{{{x^3}}}{3}} \right|_1^b }={ \frac{{{b^3}}}{3} – \frac{{{1^3}}}{3} }={ \frac{{{b^3} – 1}}{3} }={ 1.}$

Hence

${{b^3} – 1 = 3,}\;\; \Rightarrow {{b^3} = 4,}\;\; \Rightarrow {b = \sqrt{4} \approx 1.59}$

### Example 3.

Find the coordinate of the point $$a$$ that splits the area under the root function $$y = \sqrt{x}$$ on the interval $$\left[{0,4}\right]$$ into equal parts.

Solution.

Equating both areas, we get the following:

${{A_1} = {A_2},}\;\; \Rightarrow {\int\limits_0^a {\sqrt x dx} = \int\limits_a^4 {\sqrt x dx} ,}\;\; \Rightarrow {\left. {\frac{{2{x^{\frac{3}{2}}}}}{3}} \right|_0^a = \left. {\frac{{2{x^{\frac{3}{2}}}}}{3}} \right|_a^4,}\;\; \Rightarrow {\frac{{2\sqrt {{a^3}} }}{3} – 0 = \frac{{2\sqrt {{4^3}} }}{3} – \frac{{2\sqrt {{a^3}} }}{3}, }\;\;\Rightarrow {\frac{{4\sqrt {{a^3}} }}{3} = \frac{{16}}{3},}\;\; \Rightarrow {\sqrt {{a^3}} = 4,}\;\; \Rightarrow {{a^3} = 16,}\;\; \Rightarrow {a = \sqrt{{16}} \approx 2.52}$

### Example 4.

The region is bounded by the vertical lines $$x = t$$, $$x = t + \large{\frac{\pi }{2}}\normalsize$$, the $$x-$$axis, and the curve $$y = a + \cos x,$$ where $$a \ge 1.$$ Determine the value of $$t$$ at which the region has the largest area.

Solution.

The area of the region is written in the form

$\require{cancel}{A = \int\limits_t^{t + \frac{\pi }{2}} {\left( {a + \cos x} \right)dx} }={ \left. {ax + \sin x} \right|_t^{t + \frac{\pi }{2}} }={ a\left( {t + \frac{\pi }{2}} \right) + \sin \left( {t + \frac{\pi }{2}} \right) }-{ at – \sin t }={ \cancel{at} + \frac{{a\pi }}{2} + \sin \left( {t + \frac{\pi }{2}} \right) }-{ \cancel{at} – \sin t }={ \frac{{a\pi }}{2} + \sin \left( {t + \frac{\pi }{2}} \right) – \sin t.}$

Using the difference of sines identity

${\sin \alpha – \sin \beta }={ 2\cos \frac{{\alpha + \beta }}{2}\sin \frac{{\alpha – \beta }}{2},}$

we obtain

${A = \frac{{a\pi }}{2} }+{ 2\cos \frac{{t + \frac{\pi }{2} + t}}{2}\sin \frac{{\cancel{t} + \frac{\pi }{2} – \cancel{t}}}{2} }={ \frac{{a\pi }}{2} + 2\cos \left( {t + \frac{\pi }{4}} \right)\sin \frac{\pi }{4} }={ \frac{{a\pi }}{2} + 2\cos \left( {t + \frac{\pi }{4}} \right) \cdot \frac{{\sqrt 2 }}{2} }={ \frac{{a\pi }}{2} + \sqrt 2 \cos \left( {t + \frac{\pi }{4}} \right).}$

The region has the largest area when $$\cos \left( {t + \large{\frac{\pi }{4}}\normalsize} \right) = -1.$$

Solving this equation, we find

${\cos \left( {t + \frac{\pi }{4}} \right) = – 1,}\;\; \Rightarrow {t + \frac{\pi }{4} = \pi + 2\pi n,}\;\; \Rightarrow {t = \frac{{3\pi }}{4} + 2\pi n,\,n \in \mathbb{Z}.}$

### Example 5.

Find the area of the region enclosed by the curve $$y = \sqrt {x + 1}$$ and the line $$y = x + 1.$$

Solution.

It is easy to see that the curve $$y = \sqrt {x + 1}$$ and the straight line $$y = x + 1$$ intersect at the points $$x = -1,$$ $$x = 0$$ (Figure $$10$$).

Then

${A = \int\limits_{ – 1}^0 {\left[ {\sqrt {x + 1} – \left( {x + 1} \right)} \right]dx} }={ \left. {\frac{{2{{\left( {x + 1} \right)}^{\frac{3}{2}}}}}{3} – \frac{{{x^2}}}{2} – x} \right|_{ – 1}^0 }={ \left( {\frac{2}{3} – 0 – 0} \right) – \left( {0 – \frac{1}{2} + 1} \right) }={ \frac{2}{3} – \frac{1}{2} }={ \frac{1}{6}.}$

### Example 6.

Find the area of the region enclosed by the root curve $$y = \sqrt {x}$$ and the line $$y = kx,$$ where $$k \gt 0.$$

Solution.

First we find the points of intersection of both curves:

${\sqrt x = kx,}\;\; \Rightarrow {\sqrt x – kx = 0,}\;\; \Rightarrow {\sqrt x \left( {1 – k\sqrt x } \right) = 0,}\;\; \Rightarrow {{x_1} = 0,\;}\kern0pt{{x_2} = \frac{1}{{{k^2}}}.}$

Now we calculate the area $$A$$ using integration:

${A = \int\limits_0^{\frac{1}{{{k^2}}}} {\left( {\sqrt x – kx} \right)dx} }={ \left. {\frac{{2{x^{\frac{3}{2}}}}}{3} – \frac{{k{x^2}}}{2}} \right|_0^{\frac{1}{{{k^2}}}} }={ \frac{2}{3}{\left( {\frac{1}{{{k^2}}}} \right)^{\frac{3}{2}}} – \frac{k}{2}{\left( {\frac{1}{{{k^2}}}} \right)^2} }={ \frac{2}{{3{k^3}}} – \frac{1}{{2{k^3}}} }={ \left( {\frac{2}{3} – \frac{1}{2}} \right)\frac{1}{{{k^3}}} }={ \frac{1}{{6{k^3}}}.}$

### Example 7.

Find the area of the region bounded by the curve $$y = {2^x}$$ and the lines $$x = 0,$$ $$y = 2.$$

Solution.

The upper graph is $$y = 2$$ and the lower curve is $$y = {2^x}.$$ Hence, the area is

${A = \int\limits_0^1 {\left( {2 – {2^x}} \right)dx} }={ \left. {2x – \frac{{{2^x}}}{{\ln 2}}} \right|_0^1 }={ \left( {2 – \frac{2}{{\ln 2}}} \right) – \left( {0 – \frac{1}{{\ln 2}}} \right) }={ 2 – \frac{1}{{\ln 2}} \approx 0.56}$

### Example 8.

Find the area enclosed by the three petaled rose $$r = \sin 3\theta .$$

Solution.

Since each petal has the same area, we calculate the area of one petal and multiply the result by three. So we have

${A = \frac{1}{2}\int\limits_0^{\frac{\pi }{3}} {{r^2}\left( \theta \right)d\theta } }={ \frac{1}{2}\int\limits_0^{\frac{\pi }{3}} {{{\sin }^2}\left( {3\theta } \right)d\theta } }={ \frac{1}{4}\int\limits_0^{\frac{\pi }{3}} {\left[ {1 – \cos \left( {6\theta } \right)} \right]d\theta } }={ \frac{1}{4}\left. {\left[ {\theta – \frac{{\sin \left( {6\theta } \right)}}{6}} \right]} \right|_0^{\frac{\pi }{3}} }={ \frac{1}{4} \cdot \frac{\pi }{3} }={ \frac{\pi }{{12}}}$

Hence, the area of the all region is $$\large{\frac{\pi }{4}}\normalsize.$$

### Example 9.

Find the area enclosed by the cardioid $$r = 1 + \cos \theta .$$

Solution.

We can easily the area of the cardioid by integrating the polar equation in the interval $$\left[ {0,2\pi } \right].$$ This yields:

${A = \frac{1}{2}\int\limits_0^{2\pi } {{r^2}\left( \theta \right)d\theta } }={ \frac{1}{2}\int\limits_0^{2\pi } {{{\left( {1 + \cos \theta } \right)}^2}d\theta } }={ \frac{1}{2}\int\limits_0^{2\pi } {\left( {1 + 2\cos \theta + {{\cos }^2}\theta } \right)d\theta } }={ \frac{1}{2}\int\limits_0^{2\pi } {\left( {1 + 2\cos \theta + \frac{{1 + \cos 2\theta }}{2}} \right)d\theta } }={ \frac{1}{4}\int\limits_0^{2\pi } {\left( {3 + 4\cos \theta + \cos 2\theta } \right)d\theta } }={ \frac{1}{4}\left. {\left[ {3\theta + 4\sin \theta + \frac{{\sin 2\theta }}{2}} \right]} \right|_0^{2\pi } }={ \frac{1}{4} \cdot 6\pi }={ \frac{{3\pi }}{2}}$

### Example 10.

Find the area of the region bounded by the astroid $${x^{\frac{3}{2}}} + {y^{\frac{3}{2}}} = 1.$$

Solution.

We represent the equation of the astroid in parametric form:

${x\left( t \right) = {\cos ^3}t,\;}\kern0pt{y\left( t \right) = {\sin ^3}t.}$

Check by substitution:

${{x^{\frac{3}{2}}} + {y^{\frac{3}{2}}} = 1,}\;\; \Rightarrow {{\left( {{{\cos }^3}t} \right)^{\frac{3}{2}}} + {\left( {{{\sin }^3}t} \right)^{\frac{3}{2}}} = 1,}\;\; \Rightarrow {{\cos ^2}t + {\sin ^2}t = 1,}$

which is true.

Let’s now calculate the area of the region enclosed by the parametric curve.

$A = \frac{1}{2}\int\limits_0^T {\left[ {x\left( t \right)y^\prime\left( t \right) – x^\prime\left( t \right)y\left( t \right)} \right]dt} .$
${x^\prime\left( t \right) = – 3{\cos ^2}t\sin t,\;\;}\kern0pt{y^\prime\left( t \right) = 3{\sin ^2}t\cos t,}$
${A = \frac{1}{2}\int\limits_0^{2\pi } {\left[ {x\left( t \right)y’\left( t \right) – x’\left( t \right)y\left( t \right)} \right]dt} }={ \frac{1}{2}\int\limits_0^{2\pi } {\left[ {3{{\cos }^4}t\,{{\sin }^2}t + 3{{\cos }^2}t\,{{\sin }^4}t} \right]dt} }={ \frac{3}{2}\int\limits_0^{2\pi } {\left[ {{{\cos }^2}t\,{{\sin }^2}t\underbrace {\left( {{{\cos }^2}t + {{\sin }^2}t} \right)}_1} \right]dt} }={ \frac{3}{8}\int\limits_0^{2\pi } {{{\sin }^2}\left( {2t} \right)dt} }={ \frac{3}{{16}}\int\limits_0^{2\pi } {\left[ {1 – \cos \left( {4t} \right)} \right]dt} }={ \frac{3}{{16}}\left. {\left[ {t – \frac{{\sin \left( {4t} \right)}}{4}} \right]} \right|_0^{2\pi } }={ \frac{3}{{16}} \cdot 2\pi }={ \frac{{3\pi }}{8}}$